In the expansion of (5^{1/2} + 7^{1/8})^{1024 | vedclass

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(B) The general term in the expansion of $(5^{1/2} + 7^{1/8})^{1024}$ is given by $T_{r+1} = {}^{1024}C_r (5^{1/2})^{1024-r} (7^{1/8})^r$,where $0 \le

Vedclass · Accepted Answer

$129$

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(B) The general term in the expansion of $(5^{1/2} + 7^{1/8})^{1024}$ is given by $T_{r+1} = {}^{1024}C_r (5^{1/2})^{1024-r} (7^{1/8})^r$,where $0 \le r \le 1024$.This simplifies to $T_{r+1} = {}^{1024}C_r \cdot 5^{(1024-r)/2} \cdot 7^{r/8}$.For the term to be an integer,the exponents of $5$ and $7$ must be non-negative integers.Thus,$(1024-r)/2$ must be an integer,which implies $r$ must be even.Also,$r/8$ must be an integer,which implies $r$ must be a multiple of $8$.Since $r$ must be a multiple of $8$,it is automatically even.Therefore,$r$ must be of the form $8k$,where $k$ is an integer such that $0 \le 8k \le 1024$.This gives $0 \le k \le 128$.The number of possible values for $k$ is $128 - 0 + 1 = 129$.Thus,there are $129$ integral terms.

In the expansion of $(5^{1/2} + 7^{1/8})^{1024}$,the number of integral terms is

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